# Definition

If there is $\triangle ABC$ and points $D,E,F$ on the sides $BC,CA,AB$ respectively such that $\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r$, then the ratio $\frac{[DEF]}{[ABC]}=\frac{r^2-r+1}{(r+1)^2}$.

Created by Foogle and Hoogle of The Ooga Booga Tribe of The Caveman Society

# Proofs

## Proof 1

Proof by Gogobao:

We have: $\frac{DB}{BC} = \frac{r}{r+1}, \frac{DC}{BC} = \frac{1}{r+1}, \frac{EC}{AC} = \frac{r}{r+1}, \frac{EA}{AC} = \frac{1}{r+1}, \frac{FA}{BA} = \frac{r}{r+1}, \frac{FB}{BA} = \frac{1}{r+1}$

We have: $[DEF] = [ABC] - [DCE] - [FAE] - [FBD]$

$[DCE] = [ABC] \cdot \frac{DC}{CB} \cdot \frac{CE}{CA} = [ABC] \cdot \frac{r}{(r+1)^2}$

$[EAF] = [ABC] \cdot \frac{EA}{CA} \cdot \frac{FA}{BA} = [ABC] \cdot \frac{r}{(r+1)^2}$

$[FBD] = [ABC] \cdot \frac{FB}{AB} \cdot \frac{BD}{CB} = [ABC] \cdot \frac{r}{(r+1)^2}$

Therefore $[DEF] = [ABC] (1-\frac{3r}{(r+1)^2})$

So we have $\frac{[DEF]}{[ABC]} = \frac{r^2-r+1}{(r+1)^2}$

## Proof 2

Proof by franzliszt

Apply Barycentrics w.r.t. $\triangle ABC$. Then $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. We can also find that $D=\left(0,\tfrac {1}{r+1},\tfrac {r}{r+1}\right),E=\left(\tfrac {r}{r+1},0,\tfrac {1}{r+1}\right),F=\left(\tfrac {1}{r+1},\tfrac {r}{r+1},0\right)$. In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that $$\frac{[DEF]}{[ABC]}= \begin{vmatrix} 0&\tfrac {1}{r+1}&\tfrac {r}{r+1} \\ \tfrac {r}{r+1}&0&\tfrac {1}{r+1}\\ \tfrac {1}{r+1}&\tfrac {r}{r+1}&0 \end{vmatrix}=\frac{r^2-r+1}{(r+1)^2}.$$

## Proof 3

Proof by RedFireTruck (talk) 12:11, 1 February 2021 (EST):

WLOG we let $A=(0, 0)$, $B=(1, 0)$, $C=(x, y)$ for $x$, $y\in\mathbb{R}$. We then use Shoelace Forumla to get $[ABC]=\frac12|y|$. We then figure out that $D=\left(\frac{rx+1}{r+1}, \frac{ry}{r+1}\right)$, $E=\left(\frac{x}{r+1}, \frac{y}{r+1}\right)$, and $F=\left(\frac{r}{r+1}, 0\right)$ so we know that by Shoelace Formula $\frac{[DEF]}{[ABC]}=\frac{\frac12\left|\frac{r^2y-ry+y}{(r+1)^2}\right|}{\frac12|y|}=\left|\frac{r^2-r+1}{(r+1)^2}\right|$. We know that $\frac{r^2-r+1}{(r+1)^2}\ge0$ for all $r\in\mathbb{R}$ so $\left|\frac{r^2-r+1}{(r+1)^2}\right|=\frac{r^2-r+1}{(r+1)^2}$.

## Proof 4

Just use jayasharmaramankumarguptareddybavarajugopal's lemma. (Thanks to tenebrine)

## Proof 5

Proof by tigerzhang:

Consider any nondegenerate triangle $ABC$. We can apply a shear in the direction parallel to $\overline{BC}$ and a distortion in the direction perpendicular to $\overline{BC}$ to move $A$ to any point on the plane while fixing $B$ and $C$. This can define any triangle up to scaling, rotation, and orientation, so we can map points in such a way that $\triangle ABC$ is equilateral. Since we have only applied linear transformations, collinear length ratios and all area ratios are constant. Thus, we can assume that $\triangle ABC$ is equilateral. Also assume WLOG that its side length is $1$, so $AF=BD=CE=\frac{r}{r+1}$ and $AE=BF=CD=\frac{1}{r+1}$.

By the Law of Cosines, $$EF^2=AE^2+AF^2-AE \cdot AF=\left(\frac{1}{r+1}\right)^2+\left(\frac{r}{r+1}\right)^2-\frac{1}{r+1} \cdot \frac{r}{r+1}=\frac{r^2-r+1}{(r+1)^2}.$$ Since $\frac{[DEF]}{[ABC]}=\left(\frac{EF}{BC}\right)^2=EF^2$, we have the desired result.

# Application 1

## Problem

The Wooga Looga Theorem states that the solution to this problem by franzliszt:

In $\triangle ABC$ points $X,Y,Z$ are on sides $BC,CA,AB$ such that $\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac 71$. Find the ratio of $[XYZ]$ to $[ABC]$.

## Solution 1

One solution is this one by RedFireTruck (talk) 12:11, 1 February 2021 (EST):

WLOG let $A=(0, 0)$, $B=(1, 0)$, $C=(x, y)$. Then $[ABC]=\frac12|y|$ by Shoelace Theorem and $X=\left(\frac{7x+1}{8}, \frac{7y}{8}\right)$, $Y=\left(\frac{x}{8}, \frac{y}{8}\right)$, $Z=\left(\frac78, 0\right)$. Then $[XYZ]=\frac12\left|\frac{43y}{64}\right|$ by Shoelace Theorem. Therefore the answer is $\boxed{\frac{43}{64}}$.

## Solution 2

or this solution by franzliszt:

We apply Barycentric Coordinates w.r.t. $\triangle ABC$. Let $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then we find that $X=\left(0,\tfrac 18,\tfrac 78\right),Y=\left(\tfrac 78,0,\tfrac 18\right),Z=\left(\tfrac18,\tfrac78,0\right)$. In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that $$\frac{[XYZ]}{[ABC]}=\begin{vmatrix} 0&\tfrac 18&\tfrac 78\\ \tfrac 78&0&\tfrac 18\\ \tfrac18&\tfrac78&0 \end{vmatrix}=\frac{43}{64}.$$ $\blacksquare$

## Solution 3

or this solution by aaja3427:

According the the Wooga Looga Theorem, It is $\frac{49-7+1}{8^2}$. This is $\boxed{\frac{43}{64}}$

## Solution 4

or this solution by AoPS user ilovepizza2020:

We use the $\mathbf{FUNDAMENTAL~THEOREM~OF~GEOGEBRA}$ to instantly get $\boxed{\frac{43}{64}}$. (Note: You can only use this method when you are not in a contest, as this method is so overpowered that the people behind mathematics examinations decided to ban it.)

## Solution 5

or this solution by eduD_looC:

This is a perfect application of the Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemma, which results in the answer being $\boxed{\frac{43}{64}}$. A very beautiful application, which leaves graders and readers speechless. Great for math contests with proofs.

## Solution 6

or this solution by CoolJupiter:

Wow. All of your solutions are slow, compared to my sol:

By math, we have $\boxed{\frac{43}{64}}$.

~CoolJupiter ^ | EVERYONE USE THIS SOLUTION IT'S BRILLIANT ~bsu1

Yes, very BRILLIANT!


~ TheAoPSLebron

## The Best Solution

By the $1+1=\text{BREAD}$ principle, we get $\boxed{\frac{43}{64}}$. Definitely the best method. When asked, please say that OlympusHero taught you this method. Because he did.

## Easiest Solution

The answer is clearly $\boxed{\frac{43}{64}}$. We leave the proof and intermediate steps to the reader as an exercise.

## Most Practical Solution

Refer to Sun Tzu's "Art of War", page $93865081$: "The answer to the problem is $\boxed{\frac{43}{64}}$". Discovered by mutinykids, who read the entire book. If you pay him \$5 he will give more wise advice.

# Application 2

## Problem

The Wooga Looga Theorem states that the solution to this problem by Matholic:

The figure below shows a triangle ABC whose area is $72 \text{cm}^2$. If $\dfrac{AD}{DB}=\dfrac{BE}{EC}=\dfrac{CF}{FA}=\dfrac{1}{5}$, find $[DEF].$

~LaTeX-ifyed by RP3.1415

## Solution 1

is this solution by franzliszt:

We apply Barycentric Coordinates w.r.t. $\triangle ABC$. Let $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then we find that $D=(\tfrac 56,\tfrac 16,0),E=(0,\tfrac 56,\tfrac 16),F=(\tfrac16,0,\tfrac56)$. In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that$$\frac{[DEF]}{[72]}=\begin{vmatrix} \tfrac 56&\tfrac 16&0\\ 0&\tfrac 56&\tfrac 16\\ \tfrac16&0&\tfrac56 \end{vmatrix}=\frac{7}{12}$$ so $[DEF]=42$. $\blacksquare$

## Solution 2

or this solution by RedFireTruck (talk) 12:11, 1 February 2021 (EST):

By the Wooga Looga Theorem, $\frac{[DEF]}{[ABC]}=\frac{5^2-5+1}{(5+1)^2}=\frac{21}{36}=\frac{7}{12}$. We are given that $[ABC]=72$ so $[DEF]=\frac{7}{12}\cdot72=\boxed{42}$

# Application 3

## Problem

The Wooga Looga Theorem states that the solution to this problem by RedFireTruck:

Find the ratio $\frac{[GHI]}{[ABC]}$ if $\frac{AD}{DB}=\frac{BE}{EC}=\frac{CF}{FA}=\frac12$ and $\frac{DG}{GE}=\frac{EH}{HF}=\frac{FI}{ID}=1$ in the diagram below.$[asy] draw((0, 0)--(6, 0)--(4, 3)--cycle); draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle); draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle); label("A", (0, 0), SW); label("B", (6, 0), SE); label("C", (4, 3), N); label("D", (2, 0), S); label("E", (16/3, 1), NE); label("F", (8/3, 2), NW); label("G", (11/3, 1/2), SE); label("H", (4, 3/2), NE); label("I", (7/3, 1), W); [/asy]$

## Solution 1

is this solution by franzliszt:

By the Wooga Looga Theorem, $\frac{[DEF]}{[ABC]}=\frac{2^2-2+1}{(1+2)^2}=\frac 13$. Notice that $\triangle GHI$ is the medial triangle of Wooga Looga Triangle of $\triangle ABC$. So $\frac{[GHI]}{[DEF]}=\frac 14$ and $\frac{[GHI]}{[ABD]}=\frac{[DEF]}{[ABC]}\cdot\frac{[GHI]}{[DEF]}=\frac 13 \cdot \frac 14 = \frac {1}{12}$ by Chain Rule ideas.

## Solution 2

or this solution by franzliszt:

Apply Barycentrics w.r.t. $\triangle ABC$ so that $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then $D=(\tfrac 23,\tfrac 13,0),E=(0,\tfrac 23,\tfrac 13),F=(\tfrac 13,0,\tfrac 23)$. And $G=(\tfrac 13,\tfrac 12,\tfrac 16),H=(\tfrac 16,\tfrac 13,\tfrac 12),I=(\tfrac 12,\tfrac 16,\tfrac 13)$.

In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that$$\frac{[GHI]}{[ABC]}=\begin{vmatrix} \tfrac 13&\tfrac 12&\tfrac 16\\ \tfrac 16&\tfrac 13&\tfrac 12\\ \tfrac 12&\tfrac 16&\tfrac 13 \end{vmatrix}=\frac{1}{12}.$$

# Application 4

## Problem

Let $ABC$ be a triangle and $D,E,F$ be points on sides $BC,AC,$ and $AB$ respectively. We have that $\frac{BD}{DC} = 3$ and similar for the other sides. If the area of triangle $ABC$ is $16$, then what is the area of triangle $DEF$? (By ilovepizza2020)

## Solution 1

By Franzliszt

By Wooga Looga, $\frac{[DEF]}{16} = \frac{3^2-3+1}{(3+1)^2}=\frac{7}{16}$ so the answer is $\boxed7$.

## Solution 2

By franzliszt

Apply Barycentrics w.r.t. $\triangle ABC$. Then $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. We can also find that $D=(0,\tfrac 14,\tfrac 34),E=(\tfrac 34,0,\tfrac 14),F=(\tfrac 14,\tfrac 34,0)$. In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that$$\frac{[DEF]}{[ABC]}=\begin{vmatrix} 0&\tfrac 14&\tfrac 34\\ \tfrac 34&0&\tfrac 14\\ \tfrac 14&\tfrac 34&0 \end{vmatrix}=\frac{7}{16}.$$So the answer is $\boxed{7}$.

## Solution 3

A long story short, the answer must be $\boxed{7}$ by the inverse of the Inverse Wooga Looga Theorem

# Testimonials

Pogpr0 = wooga looga - Ladka13 The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. ~ilp2020

Thanks for rediscovering our theorem RedFireTruck - Foogle and Hoogle of The Ooga Booga Tribe of The Caveman Society

Franzlist is wooga looga howsopro - volkie boy

this is in fact a pretty sensible theorem. Nothing to be so excited about, though. ~DofL

The Wooga Looga Theorem is EPIC POGGERS WHOLESOME 100 KEANU CHUNGUS AMAZING SKILL THEOREM!!!!!1!!!111111 -centslordm

The Wooga Looga Theorem is amazing and can be applied to so many problems and should be taught in every school. - RedFireTruck (talk) 11:00, 1 February 2021 (EST)

The Wooga Looga Theorem is the best. -aaja3427

The Wooga Looga Theorem is needed for everything and it is great-hi..

The Wooga Looga Theorem was made by the author of the 5th Testimonial, RedFireTruck, which means they are the ooga booga tribe... proof: go to https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ and click "about". now copy and paste the aops URL. you got RedFireTruck! Great Job! now go check out his thread for post milestones, https://artofproblemsolving.com/community/c3h2319596, and give him a friend request! -FPT

This theorem has helped me with school and I am no longer failing my math class. -mchang

I, who [u]rarely[\u] edits the AoPS Wiki, has edited this to show how amazing this theorem is! The Wooga Looga theorem has actually helped me on a school test, a math competition, and more! My teacher got upset at me for not doing it the way I she taught it though - ChrisalonaLiverspur

"I can't believe AoPS books don't have this amazing theorem. If you need help with math, you can depend on caveman." ~CoolJupiter

Before the Wooga Looga Theorem, I had NO IDEA how to solve any hard geo. But, now that I've learned it, I can solve hard geo in 7 seconds ~ ilp2020 (2nd testimonial by me)

Too powerful... ~franzliszt

The Wooga Looga Theorem is so pro ~ ac142931

It is so epic and awesome that it will blow the minds of people if they saw this ~ ac142931(2nd testimonial by me)

This theorem changed my life... ~ samrocksnature

Math competitions need to ban the use of the Wooga Looga Theorem, it's just too good. ~ jasperE3

It actually can be. I never thought I'd say this, but the Wooga Looga theorem is a legit theorem. ~ jasperE3

This is franzliszt and I endorse this theorem. ~franzliszt

This theorem is too OP. ~bestzack66

This is amazing! However much it looks like a joke, it is a legitimate - and powerful - theorem. -Supernova283

Wooga Looga Theorem is extremely useful. Someone needs to make a handout on this so everyone can obtain the power of Wooga Looga ~RP3.1415

The Wooga Looga cavemen were way ahead of their time. Good job (dead) guys! -HIA2020

It's like the Ooga Booga Theorem (also OP), but better!!! - BobDBuilder321

The Wooga Looga Theorem is a special case of Routh's Theorem. So this wiki article is DEFINITELY needed. -peace

I endorse the Wooga Looga theorem for its utter usefulness and seriousness. -HamstPan38825

This is almost as OP as the Adihaya Jayasharmaramankumarguptareddybavarajugopal Lemma. Needs to be nerfed. -CoolCarsonTheRun

I ReAlLy don't get it - Senguamar HOW DARE YOU!!!!

The Wooga Looga Theorem is the base of all geometry. It is so OP that even I don't understand how to use it.

thos theroem is very prO ~ themathboi101

You know what, this is jayasharmaramankumarguptareddybavarajugopal's lemma - Ishan

If only I knew this on some contests that I had done previously... - JacobJB

The Wooga Looga Theorem is so pr0 that it needs to be nerfed. - rocketsri

"The Wooga Looga Theorem should be used in contests and should be part of geometry books." ~ Aops-g5-gethsemanea2 (talk) 21:56, 21 December 2020 (EST)

The Wooga Looga Theorem is so OP BRUH

thank for the theorem it is trivial by 1/2 ab sin(C) formula but very helpful I have used it zero times so far in competitions so it is of great use thank - bussie

I have no idea what is going on here - awesomeguy856

fuzimiao2013 waz hear

poggers theorem - awesomeming327

The Wooga Looga theorem is very OP and not to be frowned upon - Yelly314

person who invented Wooga Looga theorem is orz orz orz wooga looga theorem OP, citing it on ANY olympiad test = instant full marks -awesomeness_in_a_bun

Wooga Looga Theorem is TRASH.

HOW DARE YOU @above and @5above DISRESPECT THE WOOGA LOOGA THEOREM. THIS IS THE MOST OP THEOREM EVER AND CAN BE USED TO SOLVE EVERY PROBLEM. BECAUSE OF THIS THEOREM, I GOT A 187 ON AMC 10A AND 10B, a 665 on the AIME, AND A 420 ON THE USA(J)MO THIS YEAR. I HAVE ALSO USED THIS TO PROVE THE GOLDBACH CONJECTURE, THE TWIN PRIME CONJECTURE, EVERY SINGLE IMO PROBLEM, AND I HAVE PROVED THE RIEMANN HYPOTHESIS. I ALSO INVENTED HAND SANITIZER THAT KILLS 100% OF BACTERIA, MILK THAT IS 0% MILK, AND A TIME MACHINE WITH THIS THEOREM. THANK YOU SO MUCH @RedFireTruck FOR THIS LEGENDARY THEOREM. I WILL ALWAYS BE INDEBTED TO YOU.

Now I know how @Louis_Vuitton got so much smarter than me! :rotfl:

this is the shoddiest theorem i have ever seen

orz theorem - tigerzhang

This theorem is very useful and good, despite its slightly meme qualities. It can be used to get high scores on any mathematics competition, proof or otherwise. - dineshs

The number one theorem in the world, don't deny it. the best theorem THE BEST. you will get 7 points on the IMO by using this - kante314

I got a gold medal on the 2022 winter olympics because of this theorem ~bronzetruck2016, 2021